How to solve this partial differential equation which is a Laplace equation

  • #1
371
0

Main Question or Discussion Point

I was trying to solve this partial differential equation which arose because I wanted to find a general solution to the Laplace equation in the case f=f(x,y).

[tex]\frac{{\partial}^{2}f}{{\partial x}^{2}}+\frac{{\partial}^{2}f}{{\partial y}^{2}}=0[/tex]

Thanks in advance.
 

Answers and Replies

  • #2
432
89
The general solution would be a harmonic function, if there was a general solution for an unbounded Laplace equation. The reason why I said "would be" is because without bounds, the Laplace equation has an infinite number of solutions.

You really have to give the boundary conditions when dealing with Laplace equation, and in general for PDEs, boundary and initial conditions.
 
  • #3
1,796
53
The general solution is:

[tex]f(x,y)=g(z)+h(\overline{z}),\quad z=x+iy[/tex]

where g and h are arbitrary [itex]C^2[/itex] functions of a single variable z=x+iy. We can show this by factoring the equation:

[tex]\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right) f=\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)=\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)u=0[/tex]

where:

[tex]u=\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}[/tex]

So first solve:

[tex]u_t-iu_x=0[/tex]

which is an easy first-order equation which turns out to be u=g(x+it) then substitute that into:

[tex]f_x+if_y=g(x+it)[/tex]

another easy one to obtain so that the general solution is:

[tex]f(x,y)=g(x+it)+h(x-it)[/tex]

Note that means the general solution is not analytic since it contains [itex]\overline{z}[/itex] but analyticity is not a requirement for the solution but only differentiability.
 
Last edited:
  • #4
371
0
The general solution is:

[tex]f(x,y)=g(z)+h(\overline{z}),\quad z=x+iy[/tex]

where g and h are arbitrary [itex]C^2[/itex] functions of a single variable z=x+iy. We can show this by factoring the equation:

[tex]\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right) f=\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)\left(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\right)=\left(\frac{\partial}{\partial x}-i\frac{\partial}{\partial y}\right)u=0[/tex]

where:

[tex]u=\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y}[/tex]

So first solve:

[tex]u_t-iu_x=0[/tex]

which is an easy first-order equation which turns out to be u=g(x+it) then substitute that into:

[tex]f_x+if_y=g(x+it)[/tex]

another easy one to obtain so that the general solution is:

[tex]f(x,y)=g(x+it)+h(x-it)[/tex]

Note that means the general solution is not analytic since it contains [itex]\overline{z}[/itex] but analyticity is not a requirement for the solution but only differentiability.
Thanks.
 

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